# Some review questions.

• December 13th 2010, 08:37 PM
dmbocci
Some review questions.
1. Find the limit $lim$ as $(x,y)->(0,0)$ ${(x^2y)/(x^4+y^2)}$. Does anyone see anything that doesnt make it zero easily enough? Ive tried a few, but maybe I'm missing something.

2.Find the absolute extremes of $F(x,y)=x^2+xy R={|x|
Someone help me set this up?

3.Evaluate: Integral of region C Mdx + Ndy for $M=e^xsiny, N=e^xcosy, C: x=t-sint, y=1-cost$ from (0,0) to (2pi,0).
Looks like unit circle. I dont know how to set it up.

Thanks
• December 13th 2010, 08:50 PM
pickslides
Quote:

Originally Posted by dmbocci
1. Find the limit $lim$ as $(x,y)->(0,0)$ ${(x^2y)/(x^4+y^2)}$. Does anyone see anything that doesnt make it zero easily enough? Ive tried a few, but maybe I'm missing something.

Have you tried changing this to polar co-ordinates?
• December 14th 2010, 01:02 PM
dmbocci
I didnt think of that. But I found that $y=x^2$ makes it $(1/2)$, therefor different paths lead to different limits, so it doesnt exist. Thanks though.

Any help on the other 2 problems?
• December 14th 2010, 01:07 PM
Plato
Quote:

Originally Posted by dmbocci
1. Find the limit $lim$ as $(x,y)->(0,0)$ ${(x^2y)/(x^4+y^2)}$. Does anyone see anything that doesnt make it zero easily enough? Ive tried a few, but maybe I'm missing something.

Did you try along the path $y=x^2~?$
• December 14th 2010, 01:40 PM
dmbocci
We must have input our posts around the same time, because yea I saw what that path makes it.

So for #2, I got critical points of (0,0) (0,1) (0,-1) (0,-2) (2,1) (2,0). But do I disregard (0,-1)&(0,-2) since they are outside my boundary?
Either way, the values I got are 0,4,6. So would 0 be my minimum and if so in my answer would I say that the function has absolute min. at both (0,1) and (0,0). Or would 4 be my min at (2,0)?
And 6 would be my abs. max. at (2,1)?
Does this look right?
• December 14th 2010, 04:48 PM
SammyS
#2. Max./Min. on [-2,2]x[-1,1]
Quote:

Originally Posted by dmbocci
2.Find the absolute extremes of $F(x,y)=x^2+xy, \ R={|x|\le 2, |y|\le 1}$
Someone help me set this up?
Thanks
...

So for #2, I got critical points of (0,0) (0,1) (0,-1) (0,-2) (2,1) (2,0). But do I disregard (0,-1)&(0,-2) since they are outside my boundary?
Either way, the values I got are 0,4,6. So would 0 be my minimum and if so in my answer would I say that the function has absolute min. at both (0,1) and (0,0). Or would 4 be my min at (2,0)?
And 6 would be my abs. max. at (2,1)?
Does this look right?

How did you find the critical points?

The point (0, -1) is not outside your boundary. $\displaystyle R$ is a rectangular region with verticies (2,1), (2,-1), -2, -1), and (-2, 1).
• December 14th 2010, 06:31 PM
SammyS
Quote:

Originally Posted by dmbocci
3.Evaluate: Integral of region C Mdx + Ndy for $M=e^x\sin y, N=e^x\cos y, C: x=t-\sin t, y=1-\cos t$ from (0,0) to (2pi,0).
Looks like unit circle. I don't know how to set it up.

Thanks

$\displaystyle C$ is a contour, so I suspect it is a contour integral that's wanted.

The contour, $\displaystyle C$, is a single arch of a cycloid.

Find $\displaystyle dx$ and $\displaystyle dy$ in terms of $\displaystyle dt$ from the contour equations.

Take $\displaystyle x$ and $\displaystyle y$ from the contour equations and plug them into the equations for $\displaystyle M$ and $\displaystyle N$. The resulting integral looks like it would be messy.

Alternative: Use Green's Theorem:
$\displaystyle \oint_{C_1} Mdx + Ndy=\underset{R}{\int\int}\ \left({{\partial M}\over{\partial x}}- {{\partial N}\over{\partial y}}\right)\ dA$, where $\displaystyle C_1$ is a closed path around region, $\displaystyle R$.

Break $\displaystyle C_1$ up into path $\displaystyle C$ as defined above, and path $\displaystyle C_0$, which goes from $\displaystyle (2\pi,\ 0)$ to $\displaystyle (0,\ 0)$ along the $\displaystyle x$-axis. $\displaystyle dy = 0$ along $\displaystyle C_0$. Also, $\displaystyle M=0$ along $\displaystyle C_0$. So the integral along $\displaystyle C_0$ contributes nothing to the integral around $\displaystyle C_0$.

Therefore, $\displaystyle \oint_{C_1} Mdx + Ndy=\int_{C} Mdx + Ndy$.

Now, compute: $\displaystyle \underset{R}{\int\int}\ \left({{\partial M}\over{\partial x}}- {{\partial N}\over{\partial y}}\right)\ dy\ dx$.

This will also take some work. Good Luck.
• December 14th 2010, 08:05 PM
dmbocci
Not thinking on #2, whoops.
So idk if this is right, but now I have (-2,1)(-2,-1)(2,-1)(2,1)(-2,2)(2,-2)(1/2,-1)(-1/2,1) as my points to check. Does this look correct?
If so, my values output are 6,2,-1/2, and for (-2,2)&(2,-2)= 0 . But I would exclude those last two values?
Someone check my work?
• December 14th 2010, 09:33 PM
SammyS
#2. Max./Min. on [-2,2]x[-1,1]
Quote:

Originally Posted by dmbocci
2.Find the absolute extremes of $F(x,y)=x^2+xy,\ R=\{|x|\le2,\ |y|\le1\}$
Someone help me set this up?

Thanks

...
Quote:

Originally Posted by dmbocci
Not thinking on #2, whoops.
So idk if this is right, but now I have (-2,1)(-2,-1)(2,-1)(2,1)(-2,2)(2,-2)(1/2,-1)(-1/2,1) as my points to check. Does this look correct?
If so, my values output are 6,2,-1/2, and for (-2,2)&(2,-2)= 0 . But I would exclude those last two values?
Someone check my work?

How do we find the extrema for $\displaystyle F(x,\ y)$ over some bounded closed region, $\displaystyle R\ ?$ Yes, look for critical points.

Critical points occur at locations within the interior of $\displaystyle R,$ where $\displaystyle {{\partial F}\over {\partial x}}=0$ and $\displaystyle {{\partial F}\over {\partial y}}=0$, or where these partial derivatives do not exist. They also occur along the boundary of $\displaystyle R\$, where directional derivatives along that boundary are zero, or don't exist. Take all of that into account, and find the minimum & maximum values of $\displaystyle F$ at all of the critical points.

$\displaystyle {{\partial F}\over {\partial x}}=2x+y$. Setting this equal to zero gives: $\displaystyle y=-2x$. $\displaystyle {{\partial F}\over {\partial y}}=x$. Setting this equal to zero gives: $\displaystyle x=0$. This gives a critical point at the origin.

On the boundary $\displaystyle y=1$: $\displaystyle F(x,\ 1)=x^2+x$, so $\displaystyle {{dF}\over {dx}}=2x+1=0\ \ \implies\ \ x=-{{1}\over{2}}$. So, $\left(-{{1}\over{2}},\ 1\right)$ is a critical point. Similarly, $\left({{1}\over{2}},\ -1\right)$ is also a critical point.

On the boundary $\displaystyle x=2$: $\displaystyle F(2,\ y)=4+2y$, so $\displaystyle {{dF}\over {dy}}=2\ne0$. So there are no critical points here, except at the end points.

Critical points are thus: $(0,\ 0), \left(-{{1}\over{2}},\ 1\right),\ \left({{1}\over{2}},\ -1\right),\ (2,\ 1), \ (2,\ -1),\ (-2,\ -1),\ (-2,\ 1)$

Evaluate $\displaystyle F(x,\ y)$ at each of these points to find the extrema.
• December 14th 2010, 09:53 PM
dmbocci
I have what you have, since I have min of 2 at 2 points and max of 6 at 2 points, in my conclusion I would state all 4 points, correct?
And for x=2 I exclude that output it gives because its outside the region of our graph, correct?